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Let's say my algorithm starts with the qubit state $|0^N\rangle$. Is there a possibility to end up in a superposition where every component is made of a bitstring containing exactly an $1$ at on position and only $0$'s on every other position, i.e. $\frac{1}{\sqrt{2}^N} (|1,0^{N-1}\rangle + |01,0^{N-2}\rangle + ...+|0^{N-1},1\rangle)$

I would like to only use the standard gates/unitary transformations: $X,Y,Z,H, \text{CNOT}$

kerf
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    This is known as w-state, and you can find good answers for your question here: https://quantumcomputing.stackexchange.com/q/4350/9474. Note that the normalization factor should be $\frac{1}{\sqrt{N}}$ – Egretta.Thula Mar 24 '23 at 21:11
  • Thanks, the question is answered. – kerf Mar 24 '23 at 22:19

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It is not possible to construct the above state using the specified gates. States that can be prepared using these gates are called stabilizer states. To quote this stack exchange post, "for a stabilizer state, the 1-site reduced density matrices must be maximally mixed or pure, which [W states, like the one on your question] aren't."

user34722
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